2006 AMC 10B Problems/Problem 25
Contents
Problem
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
Solutions
Solution 1
Let be the set of the ages of Mr. Jones' children (in other words if Mr. Jones has a child who is years old). Then and . Let be the positive integer seen on the license plate. Since at least one of or is contained in , we have .
We would like to prove that , so for the sake of contradiction, let us suppose that . Then so the units digit of is . Since the number has two distinct digits, each appearing twice, another digit of must be . Since Mr. Jones can't be years old, the last two digits can't be . Therefore must be of the form , where is a digit. Since is divisible by , the sum of the digits of must be divisible by (see Divisibility rules for 9). Hence which implies . But is not divisible by , contradiction. So and is not the age of one of Mr. Jones' kids.
(We might like to check that there does, indeed, exist such a positive integer . If is not an age of one of Mr. Jones' kids, then the license plate's number must be a multiple of . Since and is the only digit multiple of that fits all the conditions of the license plate's number, the license plate's number is .)
Solution 2
Alternatively, we can see that if one of Mr. Jones' children is of the age 5, then the license plate will have to end in the digit . The license plate cannot end in the digit as is a factor of the number, so it must be even. This means that the license plate would have to have two digits, and would either be of the form or (X being the other digit in the license plate) . The condition is impossible as Mr. Jones can't be years old. If we separate the second condition, into its prime factors, we get . is prime, and therefore can't account for allowing being evenly divisible by the childrens' ages. The accounts for the 5 and one factor of 2. This leaves , but no single digit number contains all the prime factors besides 5 and 2 of the childrens' ages, thus can't be one of the childrens' ages.
Solution 3
Another way to do the problem is by the process of elimination. The only possible correct choices are the highest powers of each prime, , , , and , since indivisibility by any powers lower than these means indivisibility by a higher power of the prime (for example, indivisibility by means indivisibility by . Since the number is divisible by , it is not the answer, and the digits have to add up to . Let be the digits: since , . The only possible choices for are , , , , and . The only number that works is , but it is not divisible by .
~edited by mobius247
Solution 4
Since 5 has the "strictest" divisibility rules of the single digit numbers, we check that case first. To be divisible by 9, the other repeating digit must be 4. Seeing that 5544 is the only arrangement of 2 4's and 2 5's that is divisible by the other single digit numbers and itself is not divisible by 5, the answer is . -liu4505 ~edited by mobius247
Video Solution
https://www.youtube.com/watch?v=BigiBRo3G2c ~David
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by This is the last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.